2. The new Twinkle bulb has a standard deviation hours. A random sample of 77 light bulbs is selected from inventory. The sample mean was found to be 492 hours.

a. Find the margin of error E for a 90% confidence interval.
Round your answer to the nearest hundredths.

Margin of error = z-value (sd/√n)

Find the z-value corresponding to 90% confidence using a z-table.

sd = standard deviation
n = sample size

I hope this will help get you started.

Please someone help me? The new twinkle bulb has a standard deviation 35 hours. A random sample of 50 light bulbs is selected from inventory. The sample mean was found to be x=500 hrs. Find the margin of error for 95%, construct a 95% confidence interval for the mean life of all Twinkle bulbs. I am desperate.

To find the margin of error (E) for a 90% confidence interval, you first need to know the standard deviation of the population (sigma), which is not given in the question.

However, if the sample size is large enough (greater than 30) and the population is approximately normally distributed, we can use the sample standard deviation (s) as an estimate of the population standard deviation.

Since the standard deviation is not provided, we will assume that it refers to the sample standard deviation (s). Therefore, we can use the formula for the margin of error:

E = Z * (s / √n)

where Z is the critical value for a given level of confidence, s is the sample standard deviation, and n is the sample size.

For a 90% confidence interval, the critical value (Z) is approximately 1.645 (you can find this value in a Z-table or using statistical software).

Plug in the given values into the formula:

E = 1.645 * (s / √n)

Now, we need to find the value of s, which is the sample standard deviation. Since it's not given in the question, we cannot proceed without this information.